Solution:

$ \begin{aligned} A= & \frac{\lambda}{\varepsilon}=\left(\frac{10^{0.1 \alpha_s}-1}{10^{0.1 \alpha_p}-1}\right)^{0.5} \\\\ = & \left(\frac{10^3-1}{10^{0.1}-1}\right)^{0.5}=62.115 \\\\ k & =\frac{\Omega_p}{\Omega_s}=\frac{200}{600}=\frac{1}{3} \\\\ N & \geq \frac{\log A}{\log 1 / k} \\\\ & \geq \frac{\log 62.115}{\log 3}=3.758\\ \end{aligned}\\ $$

Rounding off $N$ to the next higher integer we get $N=4$.